# Integrate 1/cos(x)

1. Sep 14, 2009

### soopo

How can you intgertate the 1/cos(x)?

The right answer can be found in Wolfram alpha at http://www.wolframalpha.com/input/?i=1/cos(x)

My first wrong answer was ln(cos(x)).
It suggests me that you cannot use the rule, ln(x), for trigonometric functions

2. Sep 14, 2009

### Bohrok

There are several ways to integrate 1/cosx, or secx; just look on Google. You can try
$$\frac{1}{cosx}*\frac{cosx}{cosx} = \frac{cosx}{cos^2x} = \frac{cosx}{1-sin^2x}$$
and use u = sinx. Or working with secx, use the "clever substitution," as my calc book says, u = secx + tanx, du = sec2x + secxtanx dx. Then substitute u into the previous equation, get u and du together, and integrate.

3. Sep 14, 2009

### rzaidan

Hi soopo
you can integrata as follows:
sec x + tan x
∫ (1/cos x) dx=∫ sec x dx=∫ sec x ______________ dx
sec x + tan x

(sec x)^2 + sec x tan x
∫ ___________________ dx
sec x + tan x
= ln(sec x + tan x) + C since the numerator is the derivative of the denominator.
Best Regards

4. Sep 14, 2009

### soopo

Thanks Bohrok!

I use this

cosx / (1 - (sinx)^2)

I get

1 / (1-u) du = ln|1-u|

Then, putting u=sinx back to the equation

1 / (1 - sinx) + C

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5. Sep 14, 2009

### HallsofIvy

Staff Emeritus
No, you get 1/(1- u^2) du

6. Sep 14, 2009

### Preno

This may be worth remembering - the general substitution for integrating a rational function of sin, cos R(sin(x),cos(x)), which always works, is:

- t=cos(x) if R(-u,v)=-R(u,v)
- t=sin(x) if R(u,-v)=-R(u,v)
- t=tg(x) if R(-u,-v)=R(u,v)
- t=tg(x/2) in general

7. Sep 14, 2009

### soopo

Thanks for the correction.

I get

I [ 1 / (1 - u^2) = .5 ln (1+sinx) - .5 ln (1 - sinx) + C

8. Sep 14, 2009

### soopo

1. What is tg?

2. What is R(-u, v) = -R(u,v)?

9. Sep 14, 2009

### Preno

tg = tan

R(u,v) is the rational function into which you plug sin(x) and cos(x) respectively. If R is "odd with respect to sin", you substitute for cos, and vice versa. t=tan(x/2) is the general substitution which always works (but can be rather cumbersome).

10. Sep 14, 2009

### drizzle

edit, never mind :)